I need help understanding Equivalence Relations and Equivalence Classes to proceed in Topology

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I need help understanding equivalence classes and relations. I'm self-studying topology (General Topology by Willard, Introduction to Topology by Mendelson). I have read the explanations in the book, searched Wikipedia ( https://en.wikipedia.org/wiki/Equivalence_class) and this site but am still confused.

Specifically I'm confused by the statement on page 16 of Mendelson: "Two equivalence classes are either disjoint or identical.". A similar statement appears in Willard and elsewhere so I suspect the statement is correct :). Following from this is the statement: "the equivalence classes form a partition of S." (Wikipedia) It's possible I'm confused by this bit because I don't understand the preceding discussion as well as I think. An example will make clear my confusion (I hope :)).

Consider a set S consisting of various triangles, squares and circles, could I not define a relation "are both triangles"? And another: "are both circles" and "are both squares". Each of these relations is reflexive, symmetric and transitive. And I can see that these classes are disjoint AND form a partition of S.

BUT I could also consider a relation "Not a square". Two members have the relation if they are both not squares. Again, I think this is reflexive, symmetric and transitive. But, it absolutely is not the case that "Not a Square" and "Are Both circles" are "disjoint or identical". And I don't see how we can say S is partitioned with this additional equivalence class.

I feel I'm missing something basic. Must we consider all possible equivalence classes? Or are there some other rules I'm missing/misunderstanding? Perhaps a simple example will clear up my misunderstanding.

Thanks!

Best Answer

There is something fundamental going wrong in your reasoning. You are defining two different relations on $S$:

  1. $a\sim_1 b$ if and only if $a$ and $b$ are both not squares.
  2. $a\sim_2 b$ if and only if $a$ and $b$ are both circles.

However, these relations are not equivalence relations yet; there are elements in $S$ that do not relate to itself, so it is not reflexive. There are many ways to modify $\sim_1$ and $\sim_2$ such that it does define an equivalence relation. For instance:

  1. $a\sim_1 b$ if and only if $a$ and $b$ are both squares or are both not squares.
  2. $a\sim_2 b$ if and only if $a$ and $b$ are both circles or $a=b$.

Now, both relations are transitive, symmetric and reflexive and therefore equivalence relations. However, each of the relations defines its own partition of $S$.

For instance, $\sim_1$ partitions $S$ into two equivalence classes: the set $A$ that consists of all the elements of $S$ that are not squares and the set $B$ that consists of all elements of $S$ that are squares. Note that $A$ and $B$ are disjoint sets! (This is what is meant).

Similarly, $\sim_2$ partitions $S$ into many classes: the class that contains all circles and each element in $S$ that is not a circle defines its own equivalence class consisting of just itself. These classes are obviously disjoint as well.

What you did instead is compare equivalence classes defined by $\sim_1$ with classes defined by $\sim_2$.

I hope this helps.

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